Find the value of \(k\).

By considering the graph of f write down the mean of \(X\);

By considering the graph of f write down the median of \(X\);

By considering the graph of f write down the mode of \(X\).

Show that \(P(0 \leqslant X \leqslant 2) = \frac{1}{4}\).

Hence state the interquartile range of \(X\).

Calculate \(P(X \leqslant 4|X \geqslant 3)\).

attempt to equate integral to 1 (may appear later)     M1

\(k\int\limits_0^6 {\sin \left( {\frac{{\pi x}}{6}} \right){\text{d}}x = 1} \)

correct integral     A1

\(k\left[ { - \frac{6}{\pi }\cos \left( {\frac{{\pi x}}{6}} \right)} \right]_0^6 = 1\)

substituting limits     M1

\( - \frac{6}{\pi }( - 1 - 1) = \frac{1}{k}\)

\(k = \frac{\pi }{{12}}\) A1

[4 marks]

mean \( = 3\)     A1

Note:     Award A1A0A0 for three equal answers in \((0,{\text{ }}6)\).

[1 mark]

median \( = 3\)     A1

Note:     Award A1A0A0 for three equal answers in \((0,{\text{ }}6)\).

[1 mark]

mode \( = 3\)     A1

Note:     Award A1A0A0 for three equal answers in \((0,{\text{ }}6)\).

[1 mark]

\(\frac{\pi }{{12}}\int\limits_0^2 {\sin } \left( {\frac{{\pi x}}{6}} \right){\text{d}}x\)    M1

\( = \frac{\pi }{{12}}\left[ { - \frac{6}{\pi }\cos \left( {\frac{{\pi x}}{6}} \right)} \right]_0^2\)     A1

Note:     Accept without the \(\frac{\pi }{{12}}\) at this stage if it is added later.

\(\frac{\pi }{{12}}\left[ { - \frac{6}{\pi }\left( {\cos \frac{\pi }{3} - 1} \right)} \right]\)     M1

\( = \frac{1}{4}\)     AG

[4 marks]

from (c)(i) \({Q_1} = 2\)     (A1)

as the graph is symmetrical about the middle value \(x = 3 \Rightarrow {Q_3} = 4\)     (A1)

so interquartile range is

\(4 - 2\)

\( = 2\)     A1

[3 marks]

\(P(X \leqslant 4|X \geqslant 3) = \frac{{P(3 \leqslant X \leqslant 4)}}{{P(X \geqslant 3)}}\)

\( = \frac{{\frac{1}{4}}}{{\frac{1}{2}}}\)     (M1)

\( = \frac{1}{2}\)     A1

[2 marks]